Leetcode 94 Binary Tree Inorder Traversal 二叉树

二叉树的中序遍历，即左子树，根，右子树

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12    void dfs(vector<int> &ans,TreeNode *root){
13         if(!root){
14             return;
15         }
16         dfs(ans,root->left);
17         ans.push_back(root->val);
18         dfs(ans,root->right);
19     }
20     vector<int> inorderTraversal(TreeNode *root) {
21         vector<int> ans;
22         dfs(ans,root);
23         return ans;
24     }
25 };

时间： 2024-10-14 01:07:15

Leetcode 94 Binary Tree Inorder Traversal 二叉树的相关文章

leetcode 94.Binary Tree Inorder Traversal 二叉树的中序遍历

递归算法C++代码: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> in

LeetCode 94 Binary Tree Inorder Traversal (中序遍历二叉树)

Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], 1 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 题目链接:https://leetcode.com/problems/binary-t

leetCode 94.Binary Tree Inorder Traversal（二叉树中序遍历）解题思路和方法

Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? confused what "{1,#,2,3}" means? > rea

LeetCode 94 Binary Tree Inorder Traversal（二叉树的中序遍历）+（二叉树、迭代）

翻译给定一个二叉树,返回其中序遍历的节点的值. 例如: 给定二叉树为 {1, #, 2, 3} 1 2 / 3 返回 [1, 3, 2] 备注:用递归是微不足道的,你可以用迭代来完成它吗? 原文 Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3 return [1,3,2]. Note: Recursi

Leetcode 94. Binary Tree Inorder Traversal (中序遍历二叉树）

Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], 1 2 / 3 return [1,3,2]. Note: Recursive(递归) solution is trivial, could you do it iteratively(迭代)? 思路: 解法一:用递归方法很简单, (1)如果root为空,则返回NU

C++ — 迭代法遍历二叉树（LeetCode.94 - Binary Tree Inorder Traversal）

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree [1,null,2,3], 1 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 我的AC代码,递归写法比较简单,所以试试迭代: class Solution { pub

leetcode 94 Binary Tree Inorder Traversal ----- java

Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree [1,null,2,3], 1 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 求二叉树的中序遍历,要求不是用递归. 先用递归做一下,很简单. /** * Defini

94 Binary Tree Inorder Traversal(二叉树中序遍历Medium)

题目意思:二叉树中序遍历,结果存在vector<int>中解题思路:迭代迭代实现: 1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Soluti

Leetcode #94 Binary Tree Inorder Traversal

题目链接:https://leetcode.com/problems/binary-tree-inorder-traversal/ (非递归实现)二叉树的中序遍历. 1 class Solution 2 { 3 public: 4 vector<int> inorderTraversal(TreeNode *root) 5 { 6 vector<int> output; 7 if(root == NULL) 8 { 9 return output; 10 } 11 12 stack